Find the center of the circle that can be circumscribed about EFG with E(4, 4), F(4, 2), and G(8, 2
I found the answer to be (6,3) using planetcalc.com/8116/
If the center is at (h,k) and the radius is r, then you can solve
(4-h)^2+(4-k)^2 = r^2
(4-h)^2+(2-k)^2 = r^2
(8-h)^2+(2-k)^2 = r^2
or, subtracting judiciously,
(4-k)^2 = (2-k)^2
(8-h)^2 = (2-h)^2
That gives (h,k) = (5,3)
The possible answer choices:
(3, 6)
(6, 3)
(4, 2)
(4, 4)
To find the center of the circle that circumscribes triangle EFG, we can make use of the property that the perpendicular bisectors of the sides of a triangle intersect at the center of the circumcircle.
Let's start by finding the equations of the perpendicular bisectors of two sides of triangle EFG.
1. Perpendicular bisector of side EF:
- The midpoint of EF can be found by taking the average of the coordinates of E and F.
Midpoint of EF = ((x1 + x2) / 2, (y1 + y2) / 2)
= ((4 + 4) / 2, (4 + 2) / 2)
= (4, 3)
- The slope of EF can be found using the formula:
Slope of EF = (y2 - y1) / (x2 - x1)
= (2 - 4) / (4 - 4)
= -2 / 0 (undefined)
- Since the slope is undefined, the slope of the perpendicular bisector is 0.
- Using the equation of a line in point-slope form (y - y1 = m(x - x1)), where m is the slope and (x1, y1) is a point on the line, we can write the equation of the perpendicular bisector of EF as:
y - 3 = 0(x - 4)
y - 3 = 0
y = 3
2. Perpendicular bisector of side FG:
- The midpoint of FG can be found by taking the average of the coordinates of F and G.
Midpoint of FG = ((x1 + x2) / 2, (y1 + y2) / 2)
= ((4 + 8) / 2, (2 + 2) / 2)
= (6, 2)
- The slope of FG can be found using the formula:
Slope of FG = (y2 - y1) / (x2 - x1)
= (2 - 2) / (8 - 4)
= 0 / 4
= 0
- Since the slope is 0, the slope of the perpendicular bisector is undefined.
- Using the equation of a line in point-slope form (y - y1 = m(x - x1)), where m is the slope and (x1, y1) is a point on the line, we can write the equation of the perpendicular bisector of FG as:
x - 6 = 0(y - 2)
x - 6 = 0
x = 6
Now, we have the equations of two perpendicular bisectors: y = 3 and x = 6.
To find the center of the circumcircle, we need to find the intersection point of these two lines. Since one line is horizontal and the other is vertical, their intersection point will be at (6, 3). Therefore, the center of the circle that circumscribes triangle EFG is (6, 3).